> [!NOTE] Theorem > Let $u:[a,b]\to \mathbb{R}$ be a [[Real Function|real function]] that is [[Fréchet Differentiation|differentiable]] on an [[Open Real Interval|open interval]] containing $[a,b]$ such that its [[Derivative of Real Function|derivative]] $u'$ is [[Riemann integration|Darboux integrable]] on $[a,b].$ If $f$ is [[Continuous Real Function|continuous]] on the [[Image of a set under a function|image]] $u([a,b]),$ then $\int_{a}^{b} f(u(x))u'(x) \, dx = \int_{u(a)}^{u(b)} f(t) \, dt $ **Proof**: By [[Differentiablity implies Continuity]] and [[Image of Closed Real Interval under Continuous Real Function is Closed Real Interval]], $u([a,b])$ is bounded. For all $x\in u([a,b]),$ define $F(x)= \int_{u(a)}^{x} f(t) \, dx .$By [[Fundamental theorem of calculus]], $F$ is differentiable and $F'(x)=f(x)$ for each $x.$ By [[Chain rule for derivative]], $\frac{d}{dx}F(u(x))=F'(u(x))u'(x)=f(u(x))u'(x)$ The function $f\circ u$ is continuous and $u'$ is integrable so $f(u(x))u'x$ is integrable. By [[Fundamental theorem of calculus]], $\begin{aligned} \int_a^b f(u(x)) u^{\prime}(x) d x & =F(u(b))-F(u(a)) \\ & =\int_{u(a)}^{u(b)} f(t) d t-\int_{u(a)}^{u(a)} f(t) d t \\ & =\int_{u(a)}^{u(b)} f(t) d t \end{aligned}$